Given an $N$dimensional Riemannian manifold $M$, with associated Hodge $\ast$mapping $\ast$, we have the chain complex $$ \Omega^{0} {\buildrel {\text d}^\ast \over \longleftarrow} \Omega^{N} {\buildrel {\text d}^\ast \over \longleftarrow}\cdots \Omega^{N} $$ For a Kahler manifold $M$ of complex dimension $N$, with associated Hodge $\ast$mapping $\ast$, we have a second chain complex $$ \Omega^{(0,0)} {\buildrel \overline{\partial}^\ast \over \longleftarrow} \Omega^{(0,1)} {\buildrel \overline{\partial}^\ast \over \longleftarrow} \cdots \Omega^{(0,N)}. $$ What I would like to know is for which other Hermitian manifolds does this complex exist? I suppose I'm asking what condition on the Hermitian metric will give a Hodge $\ast$map for which $\overline{\partial}^\ast(\Omega^{(0,k)}) \subset \Omega^{(0,k1)}$ and $\overline{\partial}^2=0$.

$\begingroup$ Doesn't the Hodge $*$ send $\Omega^{(0, k)}$ to $\Omega^{(n, nk)}$? $\endgroup$ – Daniel Litt Mar 22 '11 at 23:34

$\begingroup$ No, actually. $*$ sends $\Omega^{(p,q)}$ to $\Omega^{(nq,np)}$. For example, on $\mathbb{C}$ with the standard metric, $* dz = *(dx+idy) = (* dx) +i (* dy) = dy  i dx = (i) (dx + i dy) = (i) dz$. $\endgroup$ – David E Speyer Mar 23 '11 at 11:56

$\begingroup$ Ah, my * is conjugate linear, and so should be conjugated. In any case, I was objecting to a (nowfixed) typo in the last line of the question. $\endgroup$ – Daniel Litt Mar 23 '11 at 17:53

$\begingroup$ In the first line, the first $\Omega^N$ should be $\Omega^1$. $\endgroup$ – Michael Albanese May 18 '12 at 15:51
Hodge $*$ and $\overline{\partial}^*$ are defined and have the named properties on any complex manifold with respect to any Hermitian metric. No Kahlerness, nor compactness, is necessary. Note that Voisin's book presents them in the chapter before she defines the Kahler condition.
John McCarthy points out that the definition of $*$ seems to vary across textbooks. Looking at the $3$ books I have available:
Voisin (Section 5.1) defines $*$ to be a $\mathbb{C}$linear map, which takes $(p,q)$forms to $(nq, np)$ forms, and for which $\overline{\partial}^* =  * \partial *$. This is the convention I have been following.
Wells (Secton V.2) defines $*$ in the same way as Voisin, but then also defines $\overline{*}$ by $\overline{*} \phi = * \overline{\phi}$, and basically doesn't mention $*$ again. He writes $\overline{\partial}^* =  \overline{*} \overline{\partial} \overline{*}$. This defines the same map as Voisin does, but generalizes more nicely to vector bundles. This is a nice expository choice; I probably would have done better to follow this in my course.
Griffiths and Harris (Section 0.6) define $*$ to be a $\mathbb{C}$antilinear map which takes $(p,q)$forms to $(np, nq)$ forms. (See a little earlier on p. 82.) This suggests that their $*$ is Wells' $\overline{*}$. I am not sure that they are exactly the same, though. Looking at $\mathbb{C}$, with the standard Kahler form, I think Wells' $\overline{*} (dz)$ should be his $* d\overline{z}$ which I think is $dy + i dx = i d \overline{z}$. That is off by $i$ from Griffiths and Harris' formula, I'm not sure why.
In any case, it appears that G and H write $*$ for a map which has the complex conjugate built in, where the others do not. They all seem to agree as to what $\overline{\partial}^*$ means.

$\begingroup$ Ok I'm confused. Let's assume that, as you say, $\ast(\Omega^{(0,k)}) \subset \Omega^{(Nk,N)}$ and look at the action of $\overline{\partial}^\ast$ on some $\omega \in \Omega^{(0,k)}$. Since $\overline{\partial}$ is zero on $\omega^{(Nk,N}$, we must have that $\overline{\partial}^\ast(\omega) = \ast (\overline{\partial} (\ast(\omega))) = 0$. Since $k$ is arbitrary, our chain complex is trivial. $\endgroup$ – John McCarthy Mar 23 '11 at 14:53

$\begingroup$ ... sorry that should be $\overline{\partial}^\ast = \ast \overline{\partial}\ast$. But the problem still holds. $\endgroup$ – John McCarthy Mar 23 '11 at 14:54

$\begingroup$ $\overline{\partial}^* =  \ast \partial \ast$. $\endgroup$ – David E Speyer Mar 23 '11 at 18:05

$\begingroup$ This may look funky, but it has the advantage that $\partial$ and $\partial^{\ast}$ are adjoint. $\endgroup$ – David E Speyer Mar 23 '11 at 18:06

$\begingroup$ You might like to look at the notes from my course last week math.lsa.umich.edu/~speyer/632/mar10.pdf $\endgroup$ – David E Speyer Mar 23 '11 at 18:10